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Proof step 8 from proof gc-heap-fill-free-h (previous step)

m0 < n, ∀ n, n0, n1, n2, nx. m0 ≤ n → # gc-heap(fill(mklparray(n), mklp(n0, n1, LP_FREE, n2)), nx, m0) = 0 ∨ gc-heap(fill(mklparray(n), mklp(n0, n1, LP_FREE, n2)), nx, m0) = ∅ ⊦ # gc-heap(fill(mklparray(n), mklp(n0, n1, LP_FREE, n2)), nx, m0) = 0, gc-heap(fill(mklparray(n), mklp(n0, n1, LP_FREE, n2)), nx, m0) = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n, n0, n1, n2, nx

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: n, n0, n1, n2, nx